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Sidana [21]
3 years ago
9

The specific heat of nickel is 0.44 J/g*⁰C. How much energy needed to change the temperature of 95.4g of nickel from 22⁰C to 32⁰

C. Is the energy absorbed or released?
Chemistry
2 answers:
valentinak56 [21]3 years ago
8 0
To determine the heat or energy needed for the process, we use the equation,
                                               H  = mcpdT
where m is the mass, cp is the specific heat and dT is the temperature difference. 
                                               H = (95.4g)(0.44 J/g°C)(32°C - 22°C)
                                                   = 419.76 J
Thus, the amount of heat that should be ABSORBED is approximately 419.76 J. 
labwork [276]3 years ago
8 0

Answer : The amount of energy needed is, 419.76 J and the energy is absorbed.

Solution :

Formula used :

Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

where,

Q = heat gained or absorbed = ?

m = mass of nickel = 95.4 g

c = specific heat of nickel = 0.44J/g^oC      

\Delta T=\text{Change in temperature}  

T_{final} = final temperature = 32^oC

T_{initial} = initial temperature = 22^oC

Now put all the given values in the above formula, we get

Q=95.4g\times 0.44J/g^oC\times (32-22)^oC

Q=419.76J

The value of Q is positive that means energy is absorbed.

Therefore, the amount of energy needed is, 419.76 J and the energy is absorbed.

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Consider the following unbalanced chemical equation. C5H12(l) + O2(g) → CO2(g) + H2O(l) If 21.9 grams of pentane (C5H12) are bur
Marina CMI [18]

Answer : The mass of water produced will be 32.78 grams.

Explanation : Given,

Mass of C_5H_{12} = 21.9 g

Molar mass of C_5H_{12} = 72.15 g/mole

Molar mass of H_2O = 18 g/mole

First we have to calculate the moles of C_5H_{12}.

\text{Moles of }C_5H_{12}=\frac{\text{Mass of }C_5H_{12}}{\text{Molar mass of }C_5H_{12}}=\frac{21.9g}{72.15g/mole}=0.3035moles

Now we have to calculate the moles of H_2O.

The balanced chemical reaction will be,

C_5H_{12}(l)+8O_2(g)\rightarrow 5CO_2(g)+6H_2O(l)

From the balanced reaction we conclude that

As, 1 mole of C_5H_{12} react to give 6 moles of H_2O

So, 0.3035 moles of C_5H_{12} react to give 0.3035\times 6=1.821 moles of H_2O

Now we have to calculate the mass of H_2O.

\text{Mass of }H_2O=\text{Moles of }H_2O\times \text{Molar mass of }H_2O

\text{Mass of }H_2O=(1.821mole)\times (18g/mole)=32.78g

Therefore, the mass of water produced will be 32.78 grams.

3 0
3 years ago
Boyle's law only works when a gas is kept at a constant temperature. Experimentally this is very tricky as changes in pressure o
Lady bird [3.3K]

The heat that creates this temperature change coming from change in the internal energy of the system as per as first law of thermodynamics.

<h3>What is Boyle's law ?</h3>

A law stating that the pressure of a given mass of an ideal gas is inversely proportional to its volume at a constant temperature.

As we know, Boyle's law only works when the gas is kept at a constant temperature

Here,

When volume of gases decreased, it means work done has occurred on the system, so the work done is used for raising internal energy of the gas and the other is released as the thermal energy.

So,

According to 1st law of thermodynamics,

we know  Q =  ΔU + W  i.e, change in internal energy and work done. So this is a reason. Changing temperature occurs.

Learn more about Internal enrgy here ;

brainly.com/question/11278589

#SPJ1

6 0
2 years ago
Guys, look at the images. I'm so stuck, and I feel as if my brain is about to get ripped out.
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Answer:

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3 0
2 years ago
19.3 g of cadmium hydroxide reacted with 15.21 g of hydrobromic acid. How many grams of water can be made?
Alecsey [184]

Answer:

m_{H_2O}=3.384gH_2O

Explanation:

Hello,

In this case, the chemical reaction is:

Cd(OH)_2+2HBr\rightarrow CdBr_2+2H_2O

Thus, we first identify the limiting reactant by computing the yielded moles of water by both of the reactants:

n_{H_2O}^{by\ Cd(OH)_2}=19.3gCd(OH)_2*\frac{1molCd(OH)_2}{146.4gCd(OH)_2}*\frac{2molH_2O}{1molCd(OH)_2}=0.264molH_2O\\\\n_{H_2O}^{by\ HBr}=15.21gHBr*\frac{1molHBr}{80.9gHBr}*\frac{2molH_2O}{2molHBr}=0.188molH_2O

In such a way, since HBr yields less water than cadmium hydroxide, we infer that HBr is the limiting one, therefore, the yielded mass of water are:

m_{H_2O}=0.188molH_2O*\frac{18gH_2O}{1molH_2O}\\ \\m_{H_2O}=3.384gH_2O

Regards.

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